RS Aggarwal Solutions Class 10 Ex 10D

SOLVE EACH OF THE FOLLOWING EQUATIONS BY USING THE METHOD OF COMPLETING THE SQUARE:

QUESTION-1:

\(x^{2}-6x+3=0\)

Solution:

\(x^{2}-6x+3=0\)

\(\Rightarrow x^{2}-6x=-3\)

\(\Rightarrow x^{2}-2\times x\times 3+ 3^{2}=-3+3^{2}\) (Adding \(3^{2}\) on both sides)

\(\Rightarrow \left ( x-3 \right )^{2}=-3+9=6\)

\(\Rightarrow x-3 =\pm \sqrt{6}\) (Taking square root on both sides)

\(\Rightarrow x-3 = \sqrt{6}\:or\:x-3=-\sqrt{6}\)

\(\Rightarrow x=3+\sqrt{6}\:or\:x=3-\sqrt{6}\)

Hence, \(3+\sqrt{6}\:and\:3-\sqrt{6}\) are the roots of the given equation.

QUESTION-2:

\(x^{2}-4x+1=0\)

Solution:

\(x^{2}-4x+1=0\)

=> \(x^{2}-4x=-1\)

=> \(x^{2}-2\times x\times 2+2^{2}=-1+2^{2}\) (Adding \(2^{2}\) on both sides)

=> \(\left ( x-2 \right )^{2}=-1+4=3\)

=> \(\Rightarrow x-2 =\pm \sqrt{3}\) (Taking square root on both sides)

\(\Rightarrow x-2 = \sqrt{3}\:or\:x-2=-\sqrt{3}\)

\(\Rightarrow x=2+\sqrt{3}\:or\:x=2-\sqrt{3}\)

Hence, \(2+\sqrt{3}\:and\:2-\sqrt{3}\) are the roots of the given equation.

QUESTION-3:

\(x^{2}+8x-2=0\)

Solution:

=> \(x^{2}+8x=2\)

=> \(x^{2}+2\times x\times 4+4^{2}=2+4^{2}\) (Adding \(4^{2}\) on both sides)

=> \(\left ( x+4 \right )^{2}=2+16=18\)

=> \(x+4=\pm \sqrt{18}=\pm 3\sqrt{2}\) (Taking square root on both sides)

=> \(x+4= 3\sqrt{2}\:or\:x+4=-3\sqrt{2}\)

=> \(x=-4+ 3\sqrt{2}\:or\:x=-4-3\sqrt{2}\)

=> Hence, \(\left ( -4+3\sqrt{2} \right )\:and\:\left ( -4-3\sqrt{2} \right )\) are the roots of the given equation.

QUESTION-4:

\(4x^{2}+4\sqrt{3}x+3=0\)

Solution:

\(4x^{2}+4\sqrt{3}x+3=0\)

=> \(4x^{2}+4\sqrt{3}x=-3\)

\(\Rightarrow \left ( 2x \right )^{2}+2\times 2x\times \sqrt{3}+\left ( \sqrt{3} \right )^{2}\) (Adding \(\left ( \sqrt{3} \right )^{2}\) on both sides)

\(\Rightarrow \left ( 2x+\sqrt{3} \right )^{2}=-3+3=0\)

\(\Rightarrow 2x+\sqrt{3}=0\)

\(\Rightarrow x=-\frac{\sqrt{3}}{2}\)

Hence, \(-\frac{\sqrt{3}}{2}\) is the repeated root of the given equation.

QUESTION-5:

\(2x^{2}+5x-3=0\)

Solution:

\(2x^{2}+5x-3=0\)

\(\Rightarrow 4x^{2}+10x-6=0\) (Multiplying both sides by 2)

\(\Rightarrow 4x^{2}+10x=6\)

\(\Rightarrow \left ( 2x \right )^{2}+2\times 2x\times \frac{5}{2}+\left ( \frac{5}{2} \right )^{2}\) [Adding \(\left ( \frac{5}{2} \right )^{2}\) on both sides]

\(\Rightarrow \left ( 2x+\frac{5}{2} \right )^{2}=6+\frac{25}{4}=\frac{24+25}{4}=\frac{49}{4}=\left ( \frac{7}{2} \right )^{2}\)

\(\Rightarrow 2x+\frac{5}{2} =\pm \frac{7}{2}\) (Taking square root on both sides)

\(\Rightarrow 2x+\frac{5}{2} = \frac{7}{2}\:or\:2x+\frac{5}{2}=-\frac{7}{2}\)

\(\Rightarrow 2x=\frac{7}{2}-\frac{5}{2}=\frac{2}{2}=1\:or\:2x=-\frac{7}{2}-\frac{5}{2}=-\frac{12}{2}=-6\)

=> x = \(\frac{1}{2}\) or x = -3

Hence, \(\frac{1}{2}\) and -3 are the roots of the given equation.

QUESTION-6:

\(3x^{2}-x-2=0\)

Solution:

\(3x^{2}-x-2=0\)

\(9x^{2}-3x-6=0\)

\(9x^{2}-3x=6\)

\(\Rightarrow \left ( 3x \right )^{2}-2\times 3x\times \frac{1}{2}+\left ( \frac{1}{2} \right )^{2}=6+\left ( \frac{1}{2} \right )^{2}\) [Adding \(\left ( \frac{1}{2} \right )^{2}\) on both sides]

\(\Rightarrow \left ( 3x-\frac{1}{2} \right )^{2}=6+\frac{1}{4}=\frac{25}{4}=\left ( \frac{5}{2} \right )^{2}\)

\(\Rightarrow 3x-\frac{1}{2}=\pm \frac{5}{2}\) (Taking square root on both sides)

\(\Rightarrow 3x-\frac{1}{2}=\frac{5}{2}\:or\:3x-\frac{1}{2}=-\frac{5}{2}\)

\(\Rightarrow 3x=\frac{5}{2}+\frac{1}{2}=\frac{6}{2}=3\:or\:3x=-\frac{5}{2}+\frac{1}{2}=-\frac{4}{2}=-2\)

=> x = 1 or x = \(-\frac{2}{3}\)

Hence, 1 and \(-\frac{2}{3}\) are the roots of the given equation.

QUESTION-7:

\(8x^{2}-14x-15=0\)

Solution:

\(8x^{2}-14x-15=0\)

\(\Rightarrow 16x^{2}-28x-30=0\) (Multiplying both sides by 2)

\(\Rightarrow 16x^{2}-28x=30\)

\(\Rightarrow \left ( 4x \right )^{2}-2\times 4x\times \frac{7}{2}+\left ( \frac{7}{2} \right )^{2}=30+\left ( \frac{7}{2} \right )^{2}\) [Adding \(\left ( \frac{7}{2} \right )^{2}\) on both sides]

\(\Rightarrow \left ( 4x-\frac{7}{2} \right )^{2}=30+\frac{49}{4}=\frac{169}{4}=\left ( \frac{13}{2} \right )^{2}\)

\(\Rightarrow 4x-\frac{7}{2}=\pm \frac{13}{2}\) (Taking square root on both sides)

\(\Rightarrow 4x-\frac{7}{2}=\frac{13}{2}\:or\:4x-\frac{7}{2}=-\frac{13}{2}\)

\(\Rightarrow 4x=\frac{13}{2}+\frac{7}{2}=\frac{20}{2}=10\:or\:4x=-\frac{13}{2}+\frac{7}{2}=-\frac{6}{2}=-3\)

=> x = \(\frac{5}{2}\) or x = \(-\frac{3}{4}\)

Hence, \(\frac{5}{2}\) and \(-\frac{3}{4}\) are the roots of the given equation.

QUESTION-8:

\(7x^{2}+3x-4=0\)

Solution:

\(7x^{2}+3x-4=0\)

\(\Rightarrow 49x^{2}+21x-28=0\) (Multiplying both sides by 7)

\(\Rightarrow 49x^{2}+21x=28\)

\(\Rightarrow \left ( 7x \right )^{2}+2\times 7x\times \frac{3}{2}+\left ( \frac{3}{2} \right )^{2}=28+\left ( \frac{3}{2} \right )^{2}\) [Adding \(\left ( \frac{3}{2} \right )^{2}\) on both sides]

\(\Rightarrow \left ( 7x+\frac{3}{2} \right )^{2}=28+\frac{9}{4}=\frac{121}{4}=\left ( \frac{11}{2} \right )^{2}\)

\(\Rightarrow 7x+\frac{3}{2}=\pm \frac{11}{2}\) (Taking square root on both sides)

\(\Rightarrow 7x+\frac{3}{2}=\frac{11}{2}\:or\:7x+\frac{3}{2}=-\frac{11}{2}\)

\(\Rightarrow 7x=\frac{11}{2}-\frac{3}{2}=\frac{8}{2}=4\:or\:7x=-\frac{11}{2}-\frac{3}{2}=-\frac{14}{2}=-7\)

=> x = \(\frac{4}{7}\) or x = -1

Hence, \(\frac{4}{7}\) and -1 are the roots of the given equation.

QUESTION-9:

\(3x^{2}-2x-1=0\)

Solution:

\(3x^{2}-2x-1=0\)

\(\Rightarrow 9x^{2}-6x-3=0\) (Multiplying both sides by 3)

\(\Rightarrow 9x^{2}-6x=3\)

\(\Rightarrow \left ( 3x \right )^{2}-2\times 1+1^{2}=3+1^{2}\) [Adding \(1^{2}\) on both sides]

\(\Rightarrow \left (3x-1 \right)^{2}=3+1=4=\left ( 2 \right )^{2}\)

\(\Rightarrow 3x-1 =\pm 2\) (Taking square root on both sides)

=> 3x – 1 = 2 or 3x – 1 = -2

=> 3x = 3 or 3x = -1

=> x = 1 or x = \(-\frac{1}{3}\)

Hence, 1 and \(-\frac{1}{3}\) are the roots of the given equation.

QUESTION-10:

\(5x^{2}-6x-2=0\)

Solution:

\(5x^{2}-6x-2=0\)

\(\Rightarrow 25x^{2}-30x-10=0\) (Multiplying both sides by 5)

\(\Rightarrow 25x^{2}-30x=10\)

\(\Rightarrow \left ( 5x \right )^{2}-2\times 5x\times3+3^{2}=10+3^{2}\) [Adding \(3^{2}\) on both sides]

\(\Rightarrow \left ( 5x-3 \right )^{2}=10+9=19\)

\(\Rightarrow 5x-3=\pm \sqrt{19}\) (Taking square root on both sides)

\(\Rightarrow 5x-3= \sqrt{19}\:or\:5x-3=-\sqrt{19}\)

\(\Rightarrow 5x=3+\sqrt{19}\:or\:5x=3-\sqrt{19}\)

\(\Rightarrow x=\frac{3+\sqrt{19}}{5}\:or\:x=\frac{3-\sqrt{19}}{5}\)

Hence, \(\frac{3+\sqrt{19}}{5}\:and\:\frac{3-\sqrt{19}}{5}\) are the roots of the given equation.

QUESTION-11:

\(\frac{2}{x^{2}}-\frac{5}{x}+2=0\)

Solution:

\(\Rightarrow \frac{2-5x+2x^{2}}{x^{2}}=0\)

\(\Rightarrow 2x^{2}-5x+2=0\)

\(\Rightarrow 4x^{2}-10x+4=0\) (Multiplying both sides by 2)

\(\Rightarrow 4x^{2}-10x=-4\)

\(\Rightarrow \left ( 2x \right )^{2}-2\times 2x\times \frac{5}{2}+\left ( \frac{5}{2} \right )^{2}=-4+\left ( \frac{5}{2} \right )^{2}\) [Adding \(\left ( \frac{5}{2} \right )^{2}\) on both sides]

\(\Rightarrow \left ( 2x-\frac{5}{2} \right )^{2}=-4+\frac{25}{4}=\frac{9}{4}=\left ( \frac{3}{2} \right )^{2}\)

\(\Rightarrow 2x-\frac{5}{2} = \pm \frac{3}{2}\) (Taking square root on both sides)

\(\Rightarrow 2x-\frac{5}{2}=\frac{3}{2}\:or\:2x-\frac{5}{2}=-\frac{3}{2}\)

\(\Rightarrow 2x = \frac{3}{2}+\frac{5}{2}=\frac{8}{2}=4\:or\:2x=-\frac{3}{2}+\frac{5}{2}=\frac{2}{2}=1\)

=> x = 2 or x = \(\frac{1}{2}\)

Hence, 2 and \(\frac{1}{2}\) are the roots of the given equation.

QUESTION-12:

\(4x^{2}+4bx-\left ( a^{2}-b^{2} \right )=0\)

Solution:

\(4x^{2}+4bx-\left ( a^{2}-b^{2} \right )=0\)

\(\Rightarrow 4x^{2}+4bx=a^{2}-b^{2}\)

\(\Rightarrow \left ( 2x \right )^{2}+2\times 2x\times b+b^{2}=a^{2}-b^{2}+b^{2}\) [Adding \(b^{2}\) on both sides]

\(\Rightarrow \left ( 2x+b \right )^{2}=a^{2}\)

\(\Rightarrow 2x+b =\pm a\) (Taking square root on both sides)

=> 2x+b = a or 2x+b = -a

=> 2x = a-b or 2x = -a-b

\(\Rightarrow x=\frac{a-b}{2}\:or\:x= -\frac{a+b}{2}\)

Hence, \(\frac{a-b}{2}\:and\:-\frac{a+b}{2}\) are the roots of the given equation.

Question-13:

\(x^{2}-\left ( \sqrt{2}+1 \right )x+\sqrt{2}=0\)

Solution:

\(x^{2}-\left ( \sqrt{2}+1 \right )x+\sqrt{2}=0\)

\(x^{2}-\left ( \sqrt{2}+1 \right )x+\sqrt{2}=-\sqrt{2} \)

\(\Rightarrow x^{2}-2\times x\times \left ( \frac{\sqrt{2}+1}{2} \right )+\left ( \frac{\sqrt{2}+1}{2} \right )^{2}=-\sqrt{2}+\left ( \frac{\sqrt{2}+1}{2} \right )^{2}\) [Adding \(\left ( \frac{\sqrt{2}+1}{2} \right )^{2}\) on both sides]

\(\Rightarrow \left [ x-\left ( \frac{\sqrt{2}+1}{2}^{2} \right ) \right ]^{2}=\frac{-4\sqrt{2}+2+1+2\sqrt{2}}{4}=\frac{2-2\sqrt{2}+1}{4}=\left ( \frac{\sqrt{2}-1}{4} \right )^{2}\)

\(\Rightarrow x-\left ( \frac{\sqrt{2}+1}{2} \right )=\pm \left ( \frac{\sqrt{2}-1}{2} \right )\) (Taking square root on both sides)

\(\Rightarrow x-\left ( \frac{\sqrt{2}+1}{2} \right )=\left ( \frac{\sqrt{2}-1}{2} \right )\:or\:x-\left ( \frac{\sqrt{2}+1}{2} \right )=-\left ( \frac{\sqrt{2}-1}{2} \right )\)

\(\Rightarrow x=\frac{\sqrt{2}+1}{2}+\frac{\sqrt{2}-1}{2}\:or\:x=\frac{\sqrt{2}+1}{2}-\frac{\sqrt{2}-1}{2}\)

\(\Rightarrow x=\frac{2\sqrt{2}}{2}=\sqrt{2}\:or\:x=\frac{2}{2}=1\)

Hence, \(\sqrt{2}\) and 1 are the roots of the given equation.

Question-14:

\(\sqrt{2}x^{2}-3x-2\sqrt{2}=0\)

Solution:

\(\sqrt{2}x^{2}-3x-2\sqrt{2}=0\)

\(\Rightarrow \sqrt{2}x^{2}-3\sqrt{2}x-4=0\) (Multiplying both sides by \(\sqrt{2}\))

\(\Rightarrow \sqrt{2}x^{2}-3\sqrt{2}x=4\)

\(\Rightarrow \left ( \sqrt{2}x \right )^{2}-2\times \sqrt{2}x\times \frac{3}{2}+\left ( \frac{3}{2} \right )^{2}=4+\left ( \frac{3}{2} \right )^{2}\) (Adding \(\left ( \frac{3}{2} \right )^{2}\) on both sides)

\(\Rightarrow \left ( \sqrt{2}x-\frac{3}{2} \right )^{2}=4+\frac{9}{4}=\frac{25}{4}=\left ( \frac{5}{2} \right )^{2}\)

\(\Rightarrow \sqrt{2}x-\frac{3}{2}=\pm \frac{5}{2}\) (Taking square root on both sides)

\(\Rightarrow \sqrt{2}x-\frac{3}{2}=\frac{5}{2}\:or\:\sqrt{2}x-\frac{3}{2}=-\frac{5}{2}\)

\(\Rightarrow \sqrt{2}x=\frac{5}{2}+\frac{3}{2}=\frac{8}{2}=4\:or\:\sqrt{2}x=-\frac{5}{2}+\frac{3}{2}=-\frac{2}{2}=-1\)

\(\Rightarrow x=\frac{4}{\sqrt{2}}=2\sqrt{2}\:or\:x=-\frac{1}{\sqrt{2}}=-\frac{\sqrt{2}}{2}\)

Hence, \(2\sqrt{2}\:and\:-\frac{\sqrt{2}}{2}\) are the roots of the given equation.

Question-15:

\(\sqrt{3}x^{2}+10x+7\sqrt{3}=0\)

Solution:

\(\sqrt{3}x^{2}+10x+7\sqrt{3}=0\)

\(\Rightarrow \sqrt{3}x^{2}+10\sqrt{3}x+21=0\) (Multiplying both sides by \(\sqrt{3}\))

\(\Rightarrow \sqrt{3}x^{2}+10\sqrt{3}x=-21\)

\(\Rightarrow \left ( \sqrt{3x} \right )^{2}+2\times \sqrt{3}x\times 5+5^{2}=-21+5^{2}\) (Adding \(5^{2}\) on both sides)

=> \(\left ( \sqrt{3}x+5 \right )^{2}=-21+25=4=2^{2}\)

=> \(\sqrt{3}x+5=\pm 2\) (Taking square root on both sides)

=> \(\sqrt{3}x+5=2\:or\:\sqrt{3}x+5=-2\)

=> \(\sqrt{3}x=-3\:or\:\sqrt{3}x=-7\)

=> x = \(-\frac{3}{\sqrt{3}}=-\sqrt{3}\:or\:x=-\frac{7}{\sqrt{3}}=-\frac{7\sqrt{3}}{3}\)

Hence, \(-3\:and\:-\frac{7\sqrt{3}}{3}\) are the roots of the given equation.

Question-16:

By using the method of completing the square, show that the equation \(2x^{2}+x+4=0\) has no real roots.

Solution:

\(2x^{2}+x+4=0\)

=> \(4x^{2}+2x+8=0\) (Multiplying both sides by 2)

=> \(4x^{2}+2x=-8\)

\(\Rightarrow \left ( 2x \right )^{2}+2\times 2x\times \frac{1}{2}+\left ( \frac{1}{2} \right )^{2}=-8+\left ( \frac{1}{2} \right )^{2}\) (Adding \(\left ( \frac{1}{2} \right )^{2}\) on both sides)

\(\Rightarrow \left ( 2x+\frac{1}{2} \right )^{2}=-8+\frac{1}{4}=-\frac{31}{4}< 0\)

But, \(\left ( 2x+\frac{1}{2} \right )^{2}\) cannot be negative for any real value of x. So, there is no real value of x satisfying the given equation.

Hence, the given equation has no real roots.


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The gas used in welding and cutting metals is _______